\(\int \frac {x^4}{(b+a x^4)^2 \sqrt [4]{b x^2+a x^4}} \, dx\) [2220]

Optimal result

Integrand size = 28, antiderivative size = 165 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\frac {\left (-b+a x^2\right ) \left (b x^2+a x^4\right )^{3/4}}{4 a b (a+b) x \left (b+a x^4\right )}+\frac {\text {RootSum}\left [a^2+a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {2 a \log (x)-2 a \log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}-\text {$\#$1}^5}\&\right ]}{32 a (a+b)} \]

[Out]

Unintegrable

Rubi [F]

\[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx \]

[In]

Int[x^4/((b + a*x^4)^2*(b*x^2 + a*x^4)^(1/4)),x]

[Out]

(2*Sqrt[x]*(b + a*x^2)^(1/4)*Defer[Subst][Defer[Int][x^8/((b + a*x^4)^(1/4)*(b + a*x^8)^2), x], x, Sqrt[x]])/(
b*x^2 + a*x^4)^(1/4)

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \int \frac {x^{7/2}}{\sqrt [4]{b+a x^2} \left (b+a x^4\right )^2} \, dx}{\sqrt [4]{b x^2+a x^4}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {x^8}{\sqrt [4]{b+a x^4} \left (b+a x^8\right )^2} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\frac {-8 b^2 x+8 a^2 x^5+\frac {1}{2} b \sqrt {x} \sqrt [4]{b+a x^2} \left (b+a x^4\right ) \text {RootSum}\left [a^2+a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {2 a \log (x)-4 a \log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+2 \log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}-\text {$\#$1}^5}\&\right ]}{32 a b (a+b) \sqrt [4]{x^2 \left (b+a x^2\right )} \left (b+a x^4\right )} \]

[In]

Integrate[x^4/((b + a*x^4)^2*(b*x^2 + a*x^4)^(1/4)),x]

[Out]

(-8*b^2*x + 8*a^2*x^5 + (b*Sqrt[x]*(b + a*x^2)^(1/4)*(b + a*x^4)*RootSum[a^2 + a*b - 2*a*#1^4 + #1^8 & , (2*a*
Log[x] - 4*a*Log[(b + a*x^2)^(1/4) - Sqrt[x]*#1] - Log[x]*#1^4 + 2*Log[(b + a*x^2)^(1/4) - Sqrt[x]*#1]*#1^4)/(
a*#1 - #1^5) & ])/2)/(32*a*b*(a + b)*(x^2*(b + a*x^2))^(1/4)*(b + a*x^4))

Maple [N/A]

Time = 0.49 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.22

method	result	size
pseudoelliptic	\(\frac {-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}+a b \right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 a \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (\textit {\_R}^{4}-a \right )}\right ) a b \,x^{5}+8 \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {3}{4}} a \,x^{2}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}+a b \right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 a \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (\textit {\_R}^{4}-a \right )}\right ) b^{2} x -8 b \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {3}{4}}}{32 \left (a \,x^{4}+b \right ) a b \left (a +b \right ) x}\)	\(201\)

[In]

int(x^4/(a*x^4+b)^2/(a*x^4+b*x^2)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/32*(-sum(1/_R*(_R^4-2*a)*ln((-_R*x+(x^2*(a*x^2+b))^(1/4))/x)/(_R^4-a),_R=RootOf(_Z^8-2*_Z^4*a+a^2+a*b))*a*b*
x^5+8*(x^2*(a*x^2+b))^(3/4)*a*x^2-sum(1/_R*(_R^4-2*a)*ln((-_R*x+(x^2*(a*x^2+b))^(1/4))/x)/(_R^4-a),_R=RootOf(_
Z^8-2*_Z^4*a+a^2+a*b))*b^2*x-8*b*(x^2*(a*x^2+b))^(3/4))/(a*x^4+b)/a/b/(a+b)/x

Fricas [F(-1)]

Timed out. \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\text {Timed out} \]

[In]

integrate(x^4/(a*x^4+b)^2/(a*x^4+b*x^2)^(1/4),x, algorithm="fricas")

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 44.72 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.16 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\int \frac {x^{4}}{\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{4} + b\right )^{2}}\, dx \]

[In]

integrate(x**4/(a*x**4+b)**2/(a*x**4+b*x**2)**(1/4),x)

[Out]

Integral(x**4/((x**2*(a*x**2 + b))**(1/4)*(a*x**4 + b)**2), x)

Maxima [N/A]

Not integrable

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.17 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {x^{4}}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} + b\right )}^{2}} \,d x } \]

[In]

integrate(x^4/(a*x^4+b)^2/(a*x^4+b*x^2)^(1/4),x, algorithm="maxima")

[Out]

integrate(x^4/((a*x^4 + b*x^2)^(1/4)*(a*x^4 + b)^2), x)

Giac [N/A]

Not integrable

Time = 0.91 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.17 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {x^{4}}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} + b\right )}^{2}} \,d x } \]

[In]

integrate(x^4/(a*x^4+b)^2/(a*x^4+b*x^2)^(1/4),x, algorithm="giac")

[Out]

integrate(x^4/((a*x^4 + b*x^2)^(1/4)*(a*x^4 + b)^2), x)

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.17 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\int \frac {x^4}{{\left (a\,x^4+b\right )}^2\,{\left (a\,x^4+b\,x^2\right )}^{1/4}} \,d x \]

[In]

int(x^4/((b + a*x^4)^2*(a*x^4 + b*x^2)^(1/4)),x)

[Out]

int(x^4/((b + a*x^4)^2*(a*x^4 + b*x^2)^(1/4)), x)